4^{\frac{1}{n}}-1=\sum _{k=0}^{\infty } \frac{\left(\frac{\ln  4}{n}\right)^k}{k!}-1 =\left(1+\frac{\ln  4}{n}+\frac{\left(\frac{\ln  4}{n}\right)^2}{2!}+\text{...}\right)-1=\frac{\ln  4}{n}+\frac{\left(\frac{\ln  4}{n}\right)^2}{2!}+\cdots